The links pretty spells out the solution already. If you really can't solve it, why don't you post your attempt here and I'll see where you are stuck on.

@Starwar Clone
–
Split the integral into \( \int_0^{x^3} - \int_0^{t^2} \). For the first integral, just apply the fundamental theorem of calculus. For the second theorem, you will get some constant number, so you can ignore it after you differentiate. Is this enough information for you?

@Starwar Clone
–
It's not necessary, but if you make the change \(t = x \cdot \text{ Sh u} \) with \(x\) constant, and use \(1 + Sh^2(u) = Ch^2(u)\) and \(dt = x \cdot \text{Ch u} \space du\), the problemm looks like easy. Tomorrow, I promise you'll have my answer, now I'm very busy, Im' leaving my house. Tomorrow, I promise....

@Starwar Clone
–
I think the rest proof 2 should be simple, I'll finish it. I promise.... but you have to say me wheteher I'm right or not... It depends also if \(x >0\), \(x = 0\) or \(x < 0\) to concrete the ending. This question is giving me questions and suggestions... For example, does \(\int_0^{t} \space dt\) make some sense?If the answer is yes, Is it \(t\)?why?... I have another proof making the change \(u = \sqrt{x^2 + t^2} + t\)... Even we can think first this question without the derivative \(\frac{d}{dx}\), I mean given \(x \) possible concrete constants, \(x =1\), \(x =2\), \(x = 100\)... and later, seeing the results, infering the last answer... And I'm also considering Pi's advice: Use fundamental theorem of Calculus... To be continued... Can you finish my 2nd proof, and I tell you what I think about your result? Come on, dare...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestHint: Fundamental Theorem of Calculus.Log in to reply

yeah but i am still not able to get the required answer will you please post the solution to this question...

Log in to reply

The links pretty spells out the solution already. If you really can't solve it, why don't you post your attempt here and I'll see where you are stuck on.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Proof 1 (attempt 1).-\[\dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \dfrac d{dx} \left( \int_{0}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt - \int_{0}^{t^2} \dfrac 1{\sqrt{x^2 + t^2}} \, dt \right) =\] Let's call \(g(x) = x^3\) and \(f(x) = \int_{0}^{x} \dfrac 1{\sqrt{x^2 + t^2}} \, dt\) Then, \[\dfrac d{dx} \int_{0}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \dfrac d{dx} f(g(x)) = \dfrac d{dx} f(x^3) = 3x^2 \cdot ...\] Sorry, I'm getting a knot for myself...

Proof 2.-\[\dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \space (*)\] Let's make the change \( t = x \cdot \text{ Sh u}\) , \(x\) constant, \(dt = x \cdot \text{Ch u} \space du\) \[ (*) = \dfrac d{dx} \int_{Sh^{-1}(x \cdot \text{ Sh}^{2}u)}^{Sh^{-1} x^2} \, \pm \space du =...\] To be continued

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

@Pi Han Goh @Guillermo Templado help will be welcomed .

Log in to reply

I agree with Pi Hang Goh, you can also make the change \(t = x\Sh(u)\) solve , the integral, and later derivate

Log in to reply

I mean \( t = x \cdot \text{Sh u}\). I have some problems with my comments..

Log in to reply

Tomorrow, I rewiewed this, and if it's not solved. I'll try to solve it, I think I'm not going to have a lot problemms.. we'll see...

Log in to reply